The only solutions are $p = 2, 3$. We can see that $2^1 = 1^3 + 1^3$ and $3^2 = 1^3 + 2^3$, so $p = 2, 3$ are valid solutions. Now assume $p > 3$. We see that $(x + y)(x^2 - xy + y^2)$ is a power of $p$, so $(x+y) \equiv 0 \bmod{p}$ and $x^2 - xy + y^2 = (x+y)^2 - 3xy \equiv -3xy \bmod{p}$. Since $p$ is not 2 or 3, we can conclude that both $x$ and $y$ are multiples of $p$, so let $x = ap$ and $y = bp$. Then we want $p^{n-3} = a^3 + b^3$. Similarly, $a$ and $b$ are multiples of $p$, and we can repeat the process forever, which is clearly impossible. (I should add that this solution is one that I posted somewhere else which is not in the forums.)